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Solve-4x-3-8xy-2-4x-0-1-8x-2-y-4y-0-2-simultaneosly-i-e-find-the-stationary-point-Thank-you-




Question Number 203525 by Mastermind last updated on 21/Jan/24
Solve:   4x^3 +8xy^2 −4x=0 -----(1)  8x^2 y−4y            =0 -----(2)  simultaneosly i.e find the stationary point      Thank you
$$\mathrm{Solve}:\: \\ $$$$\mathrm{4x}^{\mathrm{3}} +\mathrm{8xy}^{\mathrm{2}} −\mathrm{4x}=\mathrm{0}\:—–\left(\mathrm{1}\right) \\ $$$$\mathrm{8x}^{\mathrm{2}} \mathrm{y}−\mathrm{4y}\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{0}\:—–\left(\mathrm{2}\right) \\ $$$$\mathrm{simultaneosly}\:\mathrm{i}.\mathrm{e}\:\mathrm{find}\:\mathrm{the}\:\mathrm{stationary}\:\mathrm{point} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Thank}\:\mathrm{you} \\ $$
Answered by aleks041103 last updated on 21/Jan/24
(1) x=0 or x^2 +2y^2 −1=0  ⇒(1) is  { ((x=0)),((x^2 +2y^2 =1)) :}  (2) y=0 or x=±1/(√2)    ⇒ { ((4x^3 +8xy^2 −4x=0)),((8x^2 −4y=0)) :}  ⇔  { ((x=0)),((y=0)) :} or  { ((x=0)),((x=±1/(√2))) :} or  { ((x^2 +2y^2 =1)),((y=0)) :} or  { ((x^2 +2y^2 =1)),((x=±1/(√2))) :}    ⇒ { ((4x^3 +8xy^2 −4x=0)),((8x^2 −4y=0)) :}⇔(x,y)=(0,0),(±1,0),(±1/(√2),±1/2)
$$\left(\mathrm{1}\right)\:{x}=\mathrm{0}\:{or}\:{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{1}\right)\:{is}\:\begin{cases}{{x}=\mathrm{0}}\\{{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} =\mathrm{1}}\end{cases} \\ $$$$\left(\mathrm{2}\right)\:{y}=\mathrm{0}\:{or}\:{x}=\pm\mathrm{1}/\sqrt{\mathrm{2}} \\ $$$$ \\ $$$$\Rightarrow\begin{cases}{\mathrm{4}{x}^{\mathrm{3}} +\mathrm{8}{xy}^{\mathrm{2}} −\mathrm{4}{x}=\mathrm{0}}\\{\mathrm{8}{x}^{\mathrm{2}} −\mathrm{4}{y}=\mathrm{0}}\end{cases} \\ $$$$\Leftrightarrow\:\begin{cases}{{x}=\mathrm{0}}\\{{y}=\mathrm{0}}\end{cases}\:{or}\:\begin{cases}{{x}=\mathrm{0}}\\{{x}=\pm\mathrm{1}/\sqrt{\mathrm{2}}}\end{cases}\:{or}\:\begin{cases}{{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} =\mathrm{1}}\\{{y}=\mathrm{0}}\end{cases}\:{or}\:\begin{cases}{{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} =\mathrm{1}}\\{{x}=\pm\mathrm{1}/\sqrt{\mathrm{2}}}\end{cases} \\ $$$$ \\ $$$$\Rightarrow\begin{cases}{\mathrm{4}{x}^{\mathrm{3}} +\mathrm{8}{xy}^{\mathrm{2}} −\mathrm{4}{x}=\mathrm{0}}\\{\mathrm{8}{x}^{\mathrm{2}} −\mathrm{4}{y}=\mathrm{0}}\end{cases}\Leftrightarrow\left({x},{y}\right)=\left(\mathrm{0},\mathrm{0}\right),\left(\pm\mathrm{1},\mathrm{0}\right),\left(\pm\mathrm{1}/\sqrt{\mathrm{2}},\pm\mathrm{1}/\mathrm{2}\right) \\ $$
Commented by aleks041103 last updated on 21/Jan/24
Commented by Mastermind last updated on 21/Jan/24
Thank you so much man
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{man} \\ $$$$ \\ $$

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